Intro to Uniform Circular Motion Unit 4, Lesson 1 Intro to Uniform Circular Motion

Uniform Circular Motion Motion in a circle at a constant speed The direction of acceleration is towards the center of the circle! v Note the velocity is along a tangent to the circle c v₀ v v Reason #1 Reason #2 v F = ma v₀ F is towards the center of the circle Δv points toward centre Water in a bucket demo! – 10 min Δv a points toward centre so is a But wait! How is there acceleration if the speed is constant? Direction changes, so velocity changes! This special type of acceleration has a special name...

centripetal An object moving in a circle at a constant speed has acceleration: R a directed towards the centre of the circle T = Time it takes to rotate 1 cycle

A 1200 kg car rounds an unbanked (flat) curve of radius 100 m going 108 km/h. Find a) its acceleration b) the required force of friction c) the minimum coefficient of friction d) the maximum safe speed if µ=0.1 100m = 9 m/s² a) b) Fnet = ma f = ma = 1200 x 9 = 10800 N N c) N = mg f Fnet = ma d) mg

Brain Break!

A 65 kg cyclist crests a hill with a radius of 25 m. a) If his apparent weight is 300 N, find his velocity. b) How fast would he have to go to feel weightless? N 25m mg a) Fnet = ma b) N = 0 mg = ma g = a mg - N = ma 10 min

Practice: Do #1-2 on your Circular Motion Problem Set!

Practice with Uniform Circular Motion Unit 4, Lesson 2 Practice with Uniform Circular Motion

T = √(4π2R/g) = √(4π2(12)/(9.8)) = 7.0 s A 60 kg child sits on a ferris wheel of radius 12 m which rotates at 5 rpm. Find the force exerted by the seat on the child a) at the bottom b) at the top c) what period would the wheel have to have for the child to feel weightless at the top? a) Fnet = ma N – mg = ma N = mg + N = (60)(9.8) + (60)(4π2)(12)/(122) = 785 N b) mg – N = ma N = mg – ma = (60)(9.8) - (60)(4π2)(12)/(122) = 391 N c) mg = ma g = 4π2R/T2 T = √(4π2R/g) = √(4π2(12)/(9.8)) = 7.0 s 15 min mg

a) the tension in the string. b) the period of rotation. A 50 g roll of tape is flung in a horizontal circle. The 50 cm long string makes a 15° angle with horizontal. Find a) the tension in the string. b) the period of rotation. c) the velocity. Y: FT ∙ sinθ = mg a) y FT = 1.9 N FT b) X: Fnet = ma ) 15° R x FT cosθ = ma = mg 15 min T = = 0.72 s R = 0.5cos15 = 0.483 c)

Brain Break!

Practice: Do #3-6 on your Circular Motion Problem Set. 25 min

Tension and Circular Motion Unit 4, Lesson 3 Tension and Circular Motion

(Is the rod compressed or stretched?) Example: A stick rotates so that the ball attached to it (as shown) moves at a constant speed. Find the a) velocity of the ball so that the rod is compressed with a force of 3.5 N. b) tension in the rod if the ball rotates with a period of 1.5 seconds. (Is the rod compressed or stretched?) S a) Fnet = 1.5kg mg – S = 2.3 m mg v = = 4.1 m/s b) Assume rod is compressed Assume stretched Fnet = ma Fnet = ma mg + S = ma S = ma – mg = 46 N mg – S = 20 min S = mg – mg S = 14.7 – 60.5 = – 46 N (stretched)

Brain Break!

Practice: Do #7-8 on your Circular Motion Problem Set. (You can now complete the entire problem set.) 30 min